Let $SO(n)$ be the special orthogonal group. There is a self-diffeomorphism $\phi:SO(n)\to SO(n)$ taking an element $A$ to $A^{-1}$. I am interested in the induced automorphism $\phi_\ast:\pi_i(SO(n))\to \pi_i(SO(n))$. I am mostly interested in the values of $i$ in the stable range. In this range, Bott periodicity tells us that the homotopy groups are trivial, $\mathbb{Z}_2$, or $\mathbb{Z}$. The first two of these don't have any interesting automorphisms, so the question is really only interesting when the homotopy group is $\mathbb{Z}$, so the interesting version of my question is the following.

Let $i$ be in the the stable range for $SO(n)$ and suppose that $\pi_i(SO(n))\cong \mathbb{Z}$. Is $\phi_\ast(1)=1 $ or $-1$?

One computation I have done already is the case of $\pi_1(SO(2))$ where $\phi_\ast(1)=-1$. However, $i$ is not in the stable range, so this doesn't really count. Unfortunately, the first interesting version of the question involves a computation in $\pi_3(SO(5))$, and I don't really have any idea about how to think about this group.

I could imagine the answer depending on the value of $n$ and/or $i$. Sort of like how the antipodal map for spheres is homotopic to the identity in odd dimensions and not in homotopic to the identity in even dimensions.

  • 1
    $\begingroup$ I would guess that the fact that an $\mathfrak{sl}_2$ triple in the complexified Lie algebra should have a compact form that integrates to a subgroup of $G=\mathrm{SO}_5$ which will be a quotient of $SU(2)=S^3$ would be useful here perhaps? $\endgroup$
    – krm2233
    Oct 1, 2023 at 15:11
  • $\begingroup$ @krm2233, this sounds potentially useful. Could you perhaps elaborate a bit. Do you mean an $\mathfrak{sl}_2$ triple in $\mathfrak{sl}_5(\mathbb{C})$? Also, I don't really see where the $SU(2)$ you mentioned is coming from. $\endgroup$
    – Sam Ballas
    Oct 1, 2023 at 17:04
  • $\begingroup$ For H_3(G), G compact, one can look by duality and what not instead at de Rham cohomology which is the Lie algebra cohomology, and on the latter the map given by inversion is the map given by x \mapsto -x, whose action on cohomology should not be difficult to make explicit. $\endgroup$ Oct 2, 2023 at 20:00

1 Answer 1


Thinking through this a bit more, I think the following calculates the action on the inversion map on $\pi_3(\mathrm{SO}_5)$ (and indeed the method does not use anything about that group in particular, so the method should work for any compact Lie group $G$ for which $\pi_3(G) = \mathbb Z$).

For any group $G$, let $\iota_G\colon G \to G$ be the inversion map, $\iota_G(g)= g^{-1}$ for any $g \in G$. We wish to calculate the map $(\iota_{\mathrm{SO}_5})_*\colon \pi_3(\mathrm{SO}_5)\to \pi_3(\mathrm{SO}_5)$.

If $G=\mathrm{SO}_5$, then $\mathrm{Lie}(G) = \mathfrak g$ complexifies to the simple complex Lie algebra $\mathfrak{g}_{\mathbb C}$ with Weyl group of type $B_2=C_2$, i.e. the dihedral group $D_8$. You can recover $\mathfrak g$ from $\mathfrak g_{\mathbb C}$ the real Lie algebra as the real form on which the Killing form is negative definite.

The root system has 4 positive roots, and for each positive root, the root spaces $\mathfrak g_{\alpha}$ and $\mathfrak{g}_{-\alpha}$ generate a subalgebra $\mathfrak{sl}_{\alpha}$ which is isomorphic to $\mathfrak{sl}_2(\mathbb C)$. The intersection of $\mathfrak{sl}_{\alpha}$ with the compact real form of $\mathfrak{g}$ defines a Lie algebra homomorphism $i_\alpha\colon \mathfrak{su}_2 \to \mathfrak{so}_5$ with image $\mathfrak s_{\alpha}$. The homomorphism $i_\alpha$ then exponentiates to give a homomorphism of Lie groups $I_{\alpha} \colon \mathrm{SU}_2\to \mathrm{SO}_5$ which is a covering map onto its image, and hence the image $S_{\alpha}$ is isomorphic to $\mathrm{SU}_2$ or $\mathrm{SO}_3$.

Since $\mathrm{SU}_2$, as a topological space, is the $3$-sphere $S^3$, we obtain a map $(i_\alpha)_*\colon \pi_3(S^3) \to \pi_3(\mathrm{SO}_5)$. If this is nonzero $(i_{\alpha})_*$ then one can detect the action of the inversion map using it since for any group homomorphism $\phi\colon G\to H$ then $\phi\circ \iota_H = \iota_G \circ \phi$. It follows that $\phi_*\circ (\iota_{\text{SU}_2})_*= (\iota_{\mathrm{SO}_5})_*\circ\phi_*$

Now one can check that $(i_{\alpha})_*$ is nonzero by showing its image in $H_3(\mathrm{SO}_5)$ pairs nontrivially with an invariant 3-form. This 3-form on the group is is given by the 3-form $(X,Y,Z) \mapsto \langle X,[Y,Z]\rangle$ on $\mathfrak{so}_5$ where $[-.-]$ is the Lie bracket and $\langle -,-\rangle$ is the Killing form, an invariant symmetric bilinear form on $\mathrm{so}_5$.

Now if we identify $\mathrm{SU}_2$ with $\mathbb U$ the unit quaternions, $\mathbb U = \{h = a+bi+cj+dk\mid a^2+b^2+c^2+d^2=1\}$, then $h^{-1} = a-bi-cj-dk$, so that $\iota_{\mathbb U}\colon \mathbb U \to \mathbb U$ is the action of $D=\mathrm{diag}(1,-1,-1,-1)$ on $\mathbb R^4$ restricted to the $3$-sphere. Since $\det(D)=-1$, it is in the connected component containing the reflections, and hence its action on $H_3(S^3)$ is by $-1$, so that its action on $\pi_3(S^3)$ must therefore also be by $-1$.


Since $S^1$ and $S^3$ are the only (connected) spheres which are Lie groups, the above strategy doesn't generalise in any obvious way to $\pi_k(\mathrm{SO}_n)$ with $k \neq 3$. That said, there is an obvious way to relate the homotopy groups of $\mathrm{SO}_n$ to the homotopy groups of spheres, because if $$ S^n = \{ \mathbf v \in \mathbb R^{n+1}: \|v\|=1\} $$ Then it is a homogeneous space for $\mathrm{SO}_{n+1}$ with the stabilizer of a vector $\mathbf v\in S^{n}$ being $\mathrm{SO}(H)$ where $H= (\mathbb R.\mathbf v)^{\perp}$. Since $\mathrm{SO}(H)\cong \mathrm{SO}_{n}$, the long exact sequence of the fibration

$$\require{AMScd} \begin{CD} \mathrm{SO}_n @>>> \mathrm{SO}_{n+1} \\ @. @VVV \\ @. S^n \end{CD}$$

allows one to study the homotopy groups of the special orthogonal groups inductively from those of the spheres (but obviously this is not trivial to do!)

  • $\begingroup$ Thanks a lot! I have to look this over more carefully, but it seems very helpful. Do you know if there is any possibility to generalize this to even higher homotopy groups. I'm not sure this strategy would work well there since there doesn't seem to be an obvious replacement for $SU(2)\cong S^3$. I will probably wait a few days and see if anyone is able to answer the question in full generality, but if not I will go ahead and accept this answer. $\endgroup$
    – Sam Ballas
    Oct 2, 2023 at 14:44
  • $\begingroup$ I'll add a remark to the end of my answer about other homotopy groups. $\endgroup$
    – krm2233
    Oct 2, 2023 at 17:22

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